In order to find the angle between the two vectors, we follow the formula

$\displaystyle cos(\theta)=\frac{u\bullet v}{||u||*||v||}$

and solve for  

$\displaystyle \theta$

Using the vectors in the problem, we get

$\displaystyle cos(\theta)=\frac{< 4,8>\bullet < 7,-7>}{\sqrt{4^2+8^2}*\sqrt{7^2+(-7)^2}}=\frac{4(7)+8(-7)}{\sqrt{80}*\sqrt{98} }$

Simplifying we get

$\displaystyle cos(\theta)=\frac{-28}{\sqrt{7840}}$

To solve for 

$\displaystyle \theta$

we find the 

$\displaystyle cos^{-1}$

of both sides and get

$\displaystyle cos^{-1}(cos(\theta))=cos^{-1}(\frac{-28}{\sqrt{7840}})$

and find that

$\displaystyle \theta=108^o$

In order to find the angle between the two vectors, we follow the formula

$\displaystyle cos(\theta)=\frac{u\bullet v}{||u||*||v||}$

and solve for  

$\displaystyle \theta$

Using the vectors in the problem, we get

$\displaystyle cos(\theta)=\frac{< 9,-5>\bullet < 5,9>}{\sqrt{9^2+(-5)^2}*\sqrt{5^2+9^2}}=\frac{9(5)+(-5)9}{\sqrt{106}*\sqrt{106} }$

Simplifying we get

$\displaystyle cos(\theta)=0$

To solve for 

$\displaystyle \theta$

we find the 

$\displaystyle cos^{-1}$

of both sides and get

$\displaystyle cos^{-1}(cos(\theta))=cos^{-1}(0)$

and find that

$\displaystyle \theta=90^o$

In order to find the angle between the two vectors, we follow the formula

$\displaystyle cos(\theta)=\frac{u\bullet v}{||u||*||v||}$

and solve for  

$\displaystyle \theta$

Using the vectors in the problem, we get

$\displaystyle cos(\theta)=\frac{< 1,-8>\bullet < 0,6>}{\sqrt{1^2+(-8)^2}*\sqrt{0^2+6^2}}=\frac{1(0)+(-8)6}{\sqrt{65}*\sqrt{36} }$

Simplifying we get

$\displaystyle cos(\theta)=\frac{-48}{\sqrt{2340}}$

To solve for 

$\displaystyle \theta$

we find the 

$\displaystyle cos^{-1}$

of both sides and get

$\displaystyle cos^{-1}(cos(\theta))=cos^{-1}(\frac{-48}{\sqrt{2340}})$

and find that

$\displaystyle \theta=173^o$

The angle $\displaystyle \Theta$ between the vectors $\displaystyle \mathbf a = \left \langle a_1,a_2,a_3\right \rangle$ and $\displaystyle \mathbf b = \left \langle b_1,b_2,b_3\right \rangle$ is given by the following equation:

$\displaystyle \cos\Theta=\frac{\mathbf a \cdot \mathbf b}{|\mathbf a| \ |\mathbf b|}$

where $\displaystyle \mathbf a \cdot \mathbf b$ represents the cross product of the vectors $\displaystyle \mathbf a$ and $\displaystyle \mathbf b$, and $\displaystyle |\mathbf a|$ and $\displaystyle |\mathbf b|$ represent the respective magnitudes of the vectors $\displaystyle \mathbf a$ and $\displaystyle \mathbf b$.

We are given the vectors $\displaystyle \mathbf a = \left \langle 1,-1,0\right \rangle$ and $\displaystyle \mathbf b = \left \langle 3,0,0\right \rangle$. Calculate $\displaystyle \mathbf$$\displaystyle |\mathbf a|$, $\displaystyle |\mathbf b|$, and $\displaystyle \mathbf a \cdot \mathbf b$, and then substitute these results into the formula for the angle between these vectors, as shown:

$\displaystyle |\mathbf a|=\sqrt{(1)^2+(-1)^2+(0)^2}=\sqrt{2}$,

$\displaystyle |\mathbf b|=\sqrt{(3)^2+(0)^2+(0)^2}=\sqrt{9}=3$,

and

$\displaystyle \mathbf a \cdot \mathbf b = (3)(1)+(-1)(0)+(0)(0)=3$.

Hence,

$\displaystyle \cos\Theta=\frac{3}{3\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

The principal angle $\displaystyle \Theta$ for which $\displaystyle \cos\Theta=\frac{\sqrt{2}}{2}$ is $\displaystyle \Theta=45^\circ$. Hence, the angle between the vectors $\displaystyle \mathbf a = \left \langle 1,-1,0\right \rangle$ and $\displaystyle \mathbf b = \left \langle 3,0,0\right \rangle$ measures $\displaystyle 45^\circ$.

Find the angle between the gradient vector $\displaystyle \vec{\triangledown}f(0,y,z)$ and the vector $\displaystyle \vec{u}=4\hat{i}-3\hat{j}+\sqrt{3}\hat{k}$ where $\displaystyle f$ is defined as: 

$\displaystyle f(x,y,z)=\sqrt{x+1}+y+z$

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Compute the gradient by taking the partial derivative for each direction: 

$\displaystyle \vec{\triangledown}f(x,y,z)=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}$

$\displaystyle \vec{\triangledown}f(x,y,z)=\frac{1}{2\sqrt{x+1}}\hat{i}+\hat{j}+\hat{k}$

At $\displaystyle x = 0$ we have: 

$\displaystyle \vec{\triangledown}f(0,y,z)=\frac{1}{2}\hat{i}+\hat{j}+\hat{k}$

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The angle $\displaystyle \theta$ between two vectors $\displaystyle \vec{v_1}$ and $\displaystyle \vec{v_2}$ can be found using the dot product: 

$\displaystyle \vec{v_1}\cdot\vec{v_2}=\left \| \vec{v_1}\right \|\left \| \vec{v_2}\right \|cos(\theta )$

We wish to find the angle $\displaystyle \theta$ between the two vectors: 

 $\displaystyle \vec{u}=4\hat{i}-3\hat{j}+\sqrt{3}\hat{k}$

$\displaystyle \vec{\triangledown}f(0,y,z)=\frac{1}{2}\hat{i}+\hat{j}+\hat{k}$

Compute the dot product between $\displaystyle \vec{u}$ and  $\displaystyle \vec{\triangledown}f(0,y,z)$, 

$\displaystyle \vec{\triangledown}f(0,y,z)\cdot\vec{u}=\left( \frac{1}{2}\hat{i}+\hat{j}+\hat{k} \right )\cdot\left( 4\hat{i}-3\hat{j}+\sqrt{3}\hat{k}\right )=-1+\sqrt{3}$

Therefore the dot product is: 

$\displaystyle \vec{\triangledown}f(0,y,z)\cdot\vec{u}=\sqrt{3}-1$

Compute the magnitude of $\displaystyle \vec{u}$ 

$\displaystyle \left \| \vec{u}\right \|=\sqrt{16+9+3}=\sqrt{28}=2\sqrt{7}$

Compute the magnitude of  $\displaystyle \vec{\triangledown}f(0,y,z)$, 

$\displaystyle \left \| \vec{\triangledown}f(0,x,y)\right \|=\sqrt{1/4 +2}=\sqrt\frac{9}{4}=\frac{3}{2}$

Now put it all together:  

 $\displaystyle \vec{\triangledown}f(0,y,z)\cdot\vec{u} =\left \| \vec{\triangledown}f(0,x,y)\right \|\left \| \vec{u}\right \|cos(\theta )$

$\displaystyle \sqrt{3}-1 =3\sqrt{7}\cos(\theta)$

$\displaystyle \theta = \cos^{-1}\left(\frac{\sqrt{21}-\sqrt{7}}{3} \right )$